If x is positive, which of the following could be correct ordering of x2,1/x,2x
a) x2 '<' 2x '<' 1/x
b) x2 '<' 1/x '<' 2x
c) 2x '<' x2 '<' 1/x

i) None
ii) a
iii) c
iv) a and b
v) a,b,c



The most obvious thing is to say that 1/x is the smallest, but none of them start with 1/x... so I'm suspecting a trick. That means I want to try a fraction between zero and one, because I know those are "tricky."

if x = 1/2, then:
x^2 = 1/4
1/x = 2
2x = 1
so the order would be x^2 < 2x < 1/x. So (a) is possible. Eliminate i and iii. (Looks like you got this far)

if x = 3/4, then:
x^2 = 9/16
1/x = 4/3
2x = 3/2
so the order would be x^2 < 1/x < 2x. So (b) is possible. Eliminate ii. (Looks like this is where you had trouble.)

Don't forget to try more than one number - 1/2 is the obvious starting number, but you've got to make sure that other fractions wouldn't do something different...

And (c) isn't going to work b/c the only way to make x^2 smaller than 1/x is for x^2 to be between zero and one. And if that's true, x^2 will be smaller than 2x, because x gets smaller when you square it but larger when you multiply it by 2.